Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int mat[105][105];
int chico[105];
vector<pair<string, int> > res;
int main() {
  int n = 0, m = 0, g = 0, i = 0, j = 0;
  bool flag = 0;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) scanf("%d", &mat[i][j]);
  if (n <= m) {
    for (i = 0; i < n; i++) {
      g = INT_MAX;
      for (j = 0; j < m; j++) g = min(g, mat[i][j]);
      for (j = 1; j <= g; j++) res.push_back(make_pair("row", i + 1));
      for (j = 0; j < m; j++) mat[i][j] -= g;
    }
    for (i = 0; i < m; i++) {
      g = INT_MIN;
      for (j = 0; j < n; j++) g = max(g, mat[j][i]);
      for (j = 1; j <= g; j++) res.push_back(make_pair("col", i + 1));
      for (j = 0; j < m; j++) mat[j][i] -= g;
    }
    flag = 1;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (mat[i][j] != 0) {
          flag = 0;
          break;
        }
      }
    }
    if (flag) {
      printf("%d\n", (int)(res.size()));
      for (i = 0; i < res.size(); i++)
        cout << res[i].first << " " << res[i].second << endl;
    } else
      printf("-1\n");
  } else {
    for (i = 0; i < m; i++) {
      g = INT_MAX;
      for (j = 0; j < n; j++) g = min(g, mat[j][i]);
      chico[i] = g;
      for (j = 1; j <= g; j++) res.push_back(make_pair("col", i + 1));
      for (j = 0; j < n; j++) mat[j][i] -= g;
    }
    for (i = 0; i < n; i++) {
      g = INT_MIN;
      for (j = 0; j < m; j++) g = max(g, mat[i][j]);
      for (j = 1; j <= g; j++) res.push_back(make_pair("row", i + 1));
      for (j = 0; j < m; j++) mat[i][j] -= g;
    }
    flag = 1;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (mat[i][j] != 0) {
          flag = 0;
          break;
        }
      }
    }
    if (flag) {
      printf("%d\n", (int)(res.size()));
      for (i = 0; i < res.size(); i++)
        cout << res[i].first << " " << res[i].second << endl;
    } else
      printf("-1\n");
  }
  return 0;
}